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OpenStudy (anonymous):
Solve for x
120=1.5e^3x
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OpenStudy (whpalmer4):
\[120=1.5e^{3x}\]
Using the property of logs that \(\log a^n = n\log a\) take the log of both sides and solve for \(x\).
OpenStudy (whpalmer4):
(divide both sides by 1.5 first to isolate the exponential on the right)
OpenStudy (whpalmer4):
If you use the natural log, \(\ln e = 1\)...
OpenStudy (anonymous):
So is the answer 26.67 for x
OpenStudy (whpalmer4):
No, I don't think so...\(e^{80}\) is a big number!
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OpenStudy (anonymous):
x=5.5406x10^34
OpenStudy (whpalmer4):
\[120 = 1.5e^{3x}\]Divide by 1.5
\[80 = e^{3x}\]Take natural log of both sides
\[\ln 80 = \ln e^{3x} = 3x \ln e = 3x \]
\[x=\frac{\ln 80}{3} \approx 4.932\]
OpenStudy (anonymous):
ok got it
OpenStudy (anonymous):
so the x=4.932
OpenStudy (whpalmer4):
\[\ln 80 \approx 4.382\]
\[\frac{\ln 80}{3} \approx 1.461\]
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OpenStudy (anonymous):
ok I got that so then x=1.461
OpenStudy (whpalmer4):
Must have fat-fingered the calculation the first time, that's why I always check my answers!
OpenStudy (anonymous):
ok thanks lol
OpenStudy (whpalmer4):
yes, x = 1.461.
\[120 = 1.5*e^{(3*1.461) }= 1.5*e^{4.383} = 1.5*80.07 \approx 120\]
(answer is actually a smidge less than 1.461, clearly)
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